Prove subspace existence I am trying to solve the following problem:
Given the following equations system
$
\begin{equation}
\begin{cases}
20a + b + c + d + e = 0 \\
a + 20b + c + d + e = 0 \\
a + b + 20c + d + e = 0 
\end{cases}
\end{equation},
$
How to prove that its solutions set are a $\mathbb{R}^5$ subspace? How to describe it?
I am trying to solve the system, but once it demands a $\mathbb{R}^5$ subspace (I have never worked with it), I really don't know how to deal with it.
 A: The solution is an element of $\mathbb{R}^5$ since you are working in 5 unknowns and considering only real solution(s), I suppose.
To solve this, convert this into a matrix-vector equation: $$\begin{bmatrix}20&1&1&1&1\\ 1&20&1&1&1\\1&1&20&1&1\end{bmatrix}\begin{bmatrix}a\\b\\c\\d\\e\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$$ or in augmented matrix $$\left[\begin{array}{ccccc|c}20&1&1&1&1&0\\ 1&20&1&1&1&0\\1&1&20&1&1&0\end{array}\right].$$ Perform elementary row operations to get $$\left[\begin{array}{ccccc|c}1&0&0&\frac{1}{22}&\frac{1}{22}&0\\ 0&1&0&\frac{1}{22}&\frac{1}{22}&0\\0&0&1&\frac{1}{22}&\frac{1}{22}&0\end{array}\right],$$ which is equivalent to the system of equation $$\begin{cases}a+\frac{1}{22}d+\frac{1}{22}e=0\\b+\frac{1}{22}d+\frac{1}{22}e=0\\c+\frac{1}{22}d+\frac{1}{22}e=0\end{cases}.$$ Let $d,e$ be free variable. Then $a,b,c$ can be expressed in terms of $d,e$ by $$\begin{cases}a=-\frac{1}{22}d-\frac{1}{22}e\\b=-\frac{1}{22}d-\frac{1}{22}e\\c=-\frac{1}{22}d-\frac{1}{22}e\end{cases}.$$ Thus the solutions are the set $$\left\{\left.\begin{bmatrix}-\frac{1}{22}d-\frac{1}{22}e\\-\frac{1}{22}d-\frac{1}{22}e\\-\frac{1}{22}d-\frac{1}{22}e\\d\\e \end{bmatrix} \in \mathbb{R}^5\right\vert d,e\in\mathbb{R}\right\}$$ This is how you solve it: perform Gaussian-Jordan Elimination, put free variable, write out the solution(s).
Now you can directly prove that the solution set is a subspace. However, the set of solution(s) to a homogeneous system of equations (every line equal to 0) is a well-known subspace called a null space. So you don't have to to tedious work above.
Precisely, for any $m\times n$ matrix $\mathbf{A}$, the set $$\mathcal{N}(\mathbf{A})=\{\mathbf{x}\in\mathbb{R}^n\mid \mathbf{Ax=0}\}$$ is a subspace of $\mathbb{R}^n$.
To prove this, we are left to show that it is closed under addition and scalar multiplication. Let $\mathbf{x,y}\in\mathcal{N}(\mathbf{A})$ and $k\in\mathbb{R}$. We have $$\mathbf{A}(\mathbf{x}+\mathbf{y})=\mathbf{A}\mathbf{x}+\mathbf{A}\mathbf{y}=\mathbf{0}+\mathbf{0}=\mathbf{0},$$ and $$\mathbf{A}(k\mathbf{x})=k\mathbf{A}\mathbf{x}=k\mathbf{0}=\mathbf{0}.$$ Thus, we have that $\mathcal{N}(\mathbf{A})$ is a subspace of $\mathbb{R}^n$.
A: Equivalently, subtract the second equation from the first to get 19a- 19b= 0 or a= b.  With that the equations become  21a+ c+ d+ e= 0 and 2a+ 20c+ d+ e= 0.  Subtract the second of those from the first to get 19a- 19c= 0 or a= c.
a= b= c, d, and e satisfy all three equations for any c, d, and e so a solution can be written {c, c, c, d, e}.  To show that this is a subspace you need to show that it is "closed" under addition and scalar multiplication.  You can do that by showing that xu+ yv, for u and v members of that set and x and y real numbers, is also of that form.
That is, x(c1, c1, c1, d1, e1)+ y(c2, c2, c2, d2, e2)= (xc1+ yc2, xc1+ yc2, xc1+ yc2, xd1+ yd2, xe1+ yd2) which is of the form (c, c, c, d, e) with c= xc1+ yc_2, d= xd1+ yd2, and e= xe1+ ye2.
